Finding the molecular formula of a hydrocarbon

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i_love_science
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Homework Statement
The molecular formula of a gaseous hydrocarbon can be determined by combusting it completely in excess oxygen and then passing it through potassium hydroxide solution to absorb the carbon dioxide produced. In an experiment, 200 cm3 of a hydrocarbon was reacted with 1500 cm3 of oxygen. After the hydrocarbon had combusted completely, 1000 cm3 of gas remained. The volume was reduced to 200 cm3 after the gas had been passed through a solution of potassium hydroxide. All volumes were measured under the same conditions of temperature and pressure. Deduce the formula of the hydrocarbon.
Relevant Equations
Avogadro's law
Moles are proportional to volume. There is 1000 cm3 of gas (carbon dioxide gas plus water vapour) after the reaction, and 200 cm3 of water vapour after the carbon dioxide has been absorbed. Therefore, there is 800 cm3 of carbon dioxide. Since the stoichiometric ratio of carbon dioxide to the hydrocarbon is 800:200 = 4:1, then the hydrocarbon formula should have 4 carbons.

I'm stuck on the following:
The stoichiometric ratio of carbon dioxide to the hydrocarbon is 200:200 = 1:1, so there should be 2 hydrogens in the formula. This is not reasonable, could someone explain where I went wrong?

Also: does combustion usually produce water vapour or liquid water?

Thanks.
 
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i_love_science said:
There is 1000 cm3 of gas (carbon dioxide gas plus water vapour) after the reaction

What about excess oxygen?

Generally speaking we assume (unless told otherwise) that gases were measured around STP, so water was in a liquid form and didn't count into volumes.

there is 800 cm3 of carbon dioxide

That's OK.

Since the stoichiometric ratio of carbon dioxide to the hydrocarbon is 800:200 = 4:1, then the hydrocarbon formula should have 4 carbons.

Yes.

The stoichiometric ratio of carbon dioxide to the hydrocarbon is 200:200 = 1:1, so there should be 2 hydrogens in the formula. This is not reasonable, could someone explain where I went wrong?

No, as you wrote earlier it is 4:1, not sure what you do here.

Unless you mean water:hydrocarbon ratio, but then you are wrong. Think again about the identity of the excess gas.